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RSC_x

math quiz..? sin(x)=-0.5cos(x)=0.5 x=?

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i need a math help. can anone help me... sin(x)= -0.5 cos(x)= 0.5 x=? i need the x value. if anyone can write a function for this... it will be prefect thing... thanks... i was anwered this before one year this but tuday i cant..? i think someones hawe enough math knowledge here.. thanks again..

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arcsin(-1/2) = -p/6
arccos(1/2) = p/3

Of course, you can add any multiple of 2p to these answers.

If you're looking to solve the two equations simultaneously, you'll be here for a long time, as no such solution exists.

Admiral

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There is no such x that satisfies both sin(x) = -0.5 and cos(x) = 0.5. Remember:

sin2(x) + cos2(x) = 1 for all x.
(-0.5)2 + (0.5)2 = 0.5

Therefore there is no x that satisfies those equations.

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Thanks guys...
its looks like i mistaked somethings...?
also i found the some answers but the answers are not what i wanted...

i guess it's
|OC|=-0.5;
|DC|=0.5;
sin(x)=|DC|
cos(x)=|OC|


heey ... youre right my question is wrong.

i tried to gnereate my own rendering function for opengl...

now can anyone write a function for finding value of x...?

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Quote:
Original post by RSC_x
Thanks guys...
its looks like i mistaked somethings...?
also i found the some answers but the answers are not what i wanted...

i guess it's
|OC|=-0.5;
|DC|=0.5;
sin(x)=|DC|
cos(x)=|OC|


heey ... youre right my question is wrong.

i tried to gnereate my own rendering function for opengl...

now can anyone write a function for finding value of x...?


Maybe you're using a different notation, but if |x| is the absolute value of x, then isn't it impossible for |OC| = -0.5?

Also, unless I'm mistaken, that system of equations has no solution either (assuming you meant OC = 0.5 and DC = 0.5).

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Quote:
Original post by RSC_x
now can anyone write a function for finding value of x...?

For finding x given what? There is only one variable in the system, so you can only specify one constraint, if you want an answer to exist.

Quote:
|OC|=-0.5;
|DC|=0.5;
sin(x)=|DC|
cos(x)=|OC|
is overdetermined, and in this case inconsistent.

Are you looking to find the polar representation of an arbitrary Cartesian point? (0.5, -0.5) doesn't lie on the unit circle, as it is a distance 1/√2 from the origin. If you're allowed to change the radius of the circle, the new system looks like:

r.sin(x) = -0.5
r.cos(x) = 0.5

which solves to give:

r = 1/√2
x = 3p/4

Admiral

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Of course it helps if you are able to work with a right angled triangle with size of the largest side equal to one, and have some idea about normalization.


sin(7*PI/4)

[Edited by - Raghar on May 26, 2007 10:00:12 AM]

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